I know that there's a gap between z=-1 and z=1, but I'm not sure how to prove the general case that not all points can be connected by a curve in Y. So, I can see the solution, but I don't know how to argue it.

As Mark44 said, graphing them would help. For surface X, [itex]z^2= x^2+ y^2+ 1[/itex] and in cylindrical coordinates that is [itex]z^2= r^2- 1[/itex] or [itex]r^2- z^2= 1[/itex]. In the "r,z" plane, that is a hyperbola with r axis as axis of symmetry. Rotating around the z-axis, then, we have a "hyperboloid of one sheet". For the surface Y, [itex]z^2= x^2+ y^2+ 1[/itex] or [itex]z^2= r^2+ 1[/itex] which, in the 'r,z-plane" is z^2- r^2= 1, a hyperbola with the z axis as axis of symmetry. Rotating around the z-axis, we have a "hyperboloid of two sheets".

Given two points on surface X, say [itex](x_0,y_0, z_0)[/itex] and [itex](x_1,y_1,z_1)[/itex] we can draw the straight line from [itex](x_0,y_0,z_0)[/itex] to [itex](x_0,y_0,z_1)[/itex], then a straight line from that point radially inward to [itex](x_3,y_3,z_1)[/itex] where [itex]y_3/x_3= y_0/x_0[/itex] and [itex]\sqrt{x_3^2+ y_3^2}= \sqrt{x_1^2+ y_1^2}[/itex]. Finally, a circle with center at [itex](0,0,z_1)[/itex] and radius [itex]\sqrt{x_1^2+ y_1^2}[/itex] will take us to [itex](x_1,y_1,z_1)[/itex] while staying on the surface.

For surface Y, [itex]z^2= x^2+ y^2+ 1[/itex], or [itex]z= \pm\sqrt{x^2+ y^2+ 1}[/itex] which makes it clear that there is no point on the surface with [itex]-1< z< 1[/itex]. Any point [itex](x_0,y_0,z_0)[/itex] on the surface, with z> 0, can be written as [itex]z_0= \sqrt{x_0^2+ y_0^2+ 1}[/itex]. Another point on the surface is [itex](x_0, y_0, -z_0). Since no point on the surface has z between -1 and 1, it is impossible to connect those two points by a continuous curve.